大家希望能选出一个人，所有人都认识他，但同时他不认识镇上除自己以外的其他人（在此，我们默认每个人自己认识自己）。可是小镇里的人太多了，一下子大家谁也说服不了谁

第一种方法

include "stdafx.h"

#include<iostream>  
#include<vector>  
#include<string>
using namespace std;

int main()
{
	int num;
	cin >> num;
	//vector<vector<int>> heads;
	while (num!=0)
	{
		num--;
		int n;
		int m;
		cin >> n >> m;
		
		vector<int> inVec(n+1,1);
		vector<int> outVec(n + 1, 1);
		vector<int> head;

		for (int i = 0;i < m;i++)
		{
			
			int a, b;
			cin >> a >> b;
			if (a == b)
			{
				continue;
			}
			else
			{
				outVec[a]++;
				inVec[b]++;
			
			}
			

		}
		

		for (int i = 1;i < n + 1;i++)
		{
			if (inVec[i] == n &&  outVec[i] == 1)
			{
				head.push_back(i);
			}
		}
		if (head.size() == 0)
		{
			cout << 0 << endl << endl;
		}
		else
		{
			cout << head.size() << endl;
			for (int i = 0;i < head.size() - 1;i++)
			{
				cout << head[i] << " ";
			}
			cout << head[head.size() - 1] << endl;
		}
	}
	

	return 0;
}

第二种：
#include "stdafx.h"

#include<iostream>  
#include<vector>  
#include<string>
using namespace std;

//所有的人都认识他
bool getAllpeople(vector<vector<int>> relation, int people)
{
	bool result = true;
	for (int x = 1;x < relation.size();x++)
	{
		if (relation[x][people] != 1)
		{
			result = false;
			break;
		}

	}
	return result;
}
//他不认识所有的人
bool getNopeope(vector<vector<int>> relation, int people)
{
	bool result = true;
	for (int y = 1;y < relation.size();y++)
	{
		if (y == people)
		{
			continue;
		}
		if (relation[people][y] == 1)
		{
			result = false;
			break;
		}

	}
	return result;
}
int main()
{
	int num;
	cin >> num;

	while (num != 0)
	{
		num--;
		int n;
		int m;
		cin >> n >> m;
		vector<vector<int>> relation(n + 1, vector<int>(n + 1, 0));
		vector<int > head;
		for (int i = 0;i < m;i++)
		{
			int a, b;
			cin >> a >> b;
			relation[a][b] = 1;
		}
		for (int i = 1;i < relation.size();i++)
		{
			relation[i][i] = 1;
		}


		for (int people = 1;people <relation.size();people++)
		{
			if (getAllpeople(relation, people) == true && getAllpeople(relation, people) == true)
			{
				head.push_back(people);
			}
		}
		if (head.size() == 0)
		{
			cout << 0 << endl << endl;
		}
		else
		{
			cout << head.size() << endl;
			for (int j = 0;j < head.size() - 1;j++)
			{
				cout << head[j] << " ";
			}
			cout << head[head.size() - 1] << endl;

		}

	}
	

	return 0;
}

评论：第一种方法能够通过，第二种不能通过，因为第二种的占用的空间多